9. Fourier Analysis
• The eardrum responds to a sum of all the
waves arriving at a particular instant. Yet
the individual sounds are “heard.”
• Any waveform is composed of an infinite
number of simple sine waves of various
frequencies and amplitudes.
9
11. Analog to Digital Conversion (A/D)
• In converting an analog signal to an equivalent sequence
of “0’s” and “1’s”, we go through three processes:
– Sampling:
• converting continuous–time analog signals to discrete–time
analog signals.
– Quantization
• converting discrete–time analog signals to discrete–time digital
signals (finite set of amplitude levels).
– Coding
• Map each amplitude level to a binary sequence.
11
12. [1] Sampling: Mathematical Representation
• One sample of g(t) can be obtained from
• If we want to sample g(t) periodically every Ts sec then we
can repeat this process periodically
0 0 0( ) ( ) ( ) ( ) ( )sg t g t t t g t t tδ δ= × − = × −
( ) ( ) ( )
( ) ( )
( ) ( ).
s
n
s
n
s s
n
g t g t t nT
g t t nT
g nT t nT
δ
δ
δ
∞
=−∞
∞
=−∞
∞
=−∞
= −
= × −
= × −
∑
∑
∑
12
14. Sampling: Frequency-Domain Analysis (1/2)
0 1 2 3
1 2 3
( ) ( )
cos( ) cos(2 ) cos(3 )
sin( ) sin(2 ) sin(3 )
sT s
n
s s s
s s s
t t nT
a a t a t a t
b t b t b t
δ δ
ω ω ω
ω ω ω
∞
=−∞
= −
= + + + +
+ + + +
∑
K
K
2
s
sT
π
ω =
2 2
0
2 2
1 1 1
( ) ( )
s s
s
s s
T T
T
T Ts s s
a t dt t dt
T T T
δ δ
− −
= = =∫ ∫
2 2 2
0
2 2 2
2 2 2 2
( ) cos( ) ( )cos( ) ( )cos(0)
s s s
s
s s s
T T T
T s s
T T Ts s s s
a t t dt t t dt t dt
T T T T
δ ω δ ω δ
− − −
= × = = =∫ ∫ ∫
2 2 2
0
2 2 2
2 2 2
( ) sin( ) ( )sin( ) ( )sin(0) 0
s s s
s
s s s
T T T
T s s
T T Ts s s
b t t dt t t dt t dt
T T T
δ ω δ ω δ
− − −
= × = = =∫ ∫ ∫
)()()( ttgtg sTδ=
an
bn
14
15. Sampling: Frequency-Domain Analysis (2/2)
( ) ( ) ( )
1 2 2 2
( ) ( )cos( ) ( )cos(2 ) ( )cos(3 ) .
s
n
s s s
s s s s
g t g t t nT
g t g t t g t t g t t
T T T T
δ
ω ω ω
∞
=−∞
= −
= + + + +
∑
K
[ ] [ ]
[ ]
1 1 1
( ) ( ) ( ) ( ) ( 2 ) ( 2 )
1
( 3 ) ( 3 ) .
1
( )
s s s s
s s s
s s
s
s
ns
G G G G G G
T T T
G G
T
G n
T
ω ω ω ω ω ω ω ω ω ω
ω ω ω ω
ω ω
∞
=−∞
= + − + + + − + +
+ − + + +
= −∑
K
15
18. Sampling Theorem
• A baseband signal whose spectrum is band-
limited to B Hz can be reconstructed exactly
(without any error) from its samples taken
uniformly at a rate fs ≥ 2B.
• fs ≥ 2B is called Nyquist Criterion of sampling.
• fs = 2B is called the Nyquist rate of sampling.
• Does Sampling Theorem Make Sense?
18
19. Reconstructing the Signal: Time-Domain Prespective
)()( ωGtg ↔
LPF
H(ω) = Ts rect(f/fs)
)()( ωGtg ↔
−
=
∗
−=
∗
−=
∗=
⋅=
⇔
∑
∑
∑
∞
−∞=
∞
−∞=
∞
−∞=
s
s
n
s
sn
ss
s
n
s
s
s
s
s
s
s
T
nTt
nTgtg
T
t
nTtnTgtg
tnTttgttgtg
ω
TGG
t
ω
T
π
πδ
ω
δ
ω
ω
ωω
ω
ω
sinc)()(
sinc)()()(
2
sinc)()(
2
sinc)()(
rect)()(
2
sincrect
19
21. Aliasing
• Sampling a signal at a rate less that the Nyquist rate results in
Aliasing.
• In aliasing, the higher frequency components take the identity of
lower frequencies.
• Real life Example: Sampling a rotating wheel.
21
22. Time Division Multiplexing (TDM)
• Multiplexing: The process of
sending two or more signals
together
– FDM: Sending them together at the
same time over different bands
using carrier modulation (AM & FM
broadcasting)
– TDM: Sending them together over
the same band by sampling the
signals and sending the samples at
different time instants (interleaved).
22
23. How to Transmit the Samples?
• Analog Pulse Modulation:
Use the samples to modulate a carrier of
pulses
– Pulse Amplitude Modulation (PAM)
– Pulse Width Modulation (PWM)
– Pulse Position Modulation (PPM)
• Pulse Code Modulation (PCM)
– Quantization of samples
– Coding
23
27. [2] Quantization
• Analog samples with an amplitude that may take value in
a specific range are converted to a digital samples with
an amplitude that takes one of a specific pre–defined set
of values.
• The range of possible values of the analog samples is
divide into L levels. L is usually taken to be a power of 2
(L = 2n
).
• The center value of each level is assigned to any sample
that falls in that quantization interval.
• For almost all samples, the quantized samples will differ
from the original samples by a small amount, called the
quantization error.
27
32. • We want to scan and send a black-and-white image of
height 11 inches and width 8.5 inches (Letter size paper).
The resolution of the scanner is 600×600 dots per inch
square. The picture will be quantized using 256 levels.
Find the size of the scanned image and the time it takes
to transmit the image using a modem of speed 56 kbps.
– Size of image =
11(in)×8.5(in)×600×600(samples/in2
)×8bits/sample
= 269280000 bits = 269 Mbits
– Time to transmit = 269280000 / 56,000 = 4808 sec = 80 min
32
33. How would 0’s and 1’s be transmitted?
• The simplest form is to send a +ve pulse for a
“1” and a –ve pulse for a “0”.
• Transmitting the message g(t) would translate
into sending a a long sequence of +ve and –ve
pulses.
33
34. Advantages of Digital Communications
• Rugged: Can withstand channel noise and
distortion much better.
• Use of repeaters (travels as far as needed).
• Use of TDM
• Can be encrypted (Security and Privacy)
• Can be encoded for error correction
(reliability).
• Easy to process, store and search.
34
35. Nyquist Theorem for Transmission
• Note that the larger the transmission rate
(pulses/sec) the narrower the pulse, the wider
its spectrum, the higher the channel
bandwidth required for transmission.
• The minimum theoretical bandwidth
required to transmit R pulses/sec is R/2 Hz.
(To be demonstrated later)
35
36. • A signal m(t) band-limited to 3 kHz is sampled
at a rate 33.33% higher than the Nyquist rate,
quantized and coded. The maximum
acceptable quantization error is 0.5% of mp.
Find the minimum bandwidth required for
transmission? How is that compared to SSB?
• Ans: 32 kHz.
36
37. TDM Revisited
• Time axis is divided into frames. Frame rate is
determined by sampling rate.
• Each frame is divided into slots.
• Each user is assigned a slot (periodically in each frame).
• A user uses the full bandwidth during his slot.
• The transmission rate of the multiplexed channel is the
sum of the rates of individual channels plus the control
bits.
• Can be used with digital signals only.
37
38. TDM in Telephony (T1 & E1 Systems)
• T1:
– Introduced in 1960s
– North America and Japan
• E1 system (Europe): 30 voice channels + 2 syn channels
2
24
1
MUXMUX
1
2
24
24 b1 2 . . .b2322
frame
24
...
...
38
39. T1 System
• Multiplexes 24 voice channels
• Voice bandwidth is approximately 3.4 kHz
• Nyquist rate of sampling = 6800 samples/sec
• Actual sampling rate = 8000 samples/sec
• 8 bits/sample (256 levels)
• Frame duration = 1/8000 = 125 µsec
• Number of bits/frame = 24×8+1=193
• Bit duration = 0.647 µsec
• Transmission rate:
(24×8+1) bits/frame × 8000 frames/sec = 1.544 Mbps
39
40. Quantization Noise
• The quantization error is assumed to be uniformly
distributed over the range (-∆ν/2,∆ν/2).
( ) ( ) ( ) ( )
( )
/2/2 3
2
/2 / 2
3 3 3 3
2
1 1
3
/ 2 / 21 1
3 3 24 24
12
vv
q
v q v
q
P q dq
v v
v v v v
v v
v
∆∆
−∆ =−∆
= =
∆ ∆
∆ −∆ ∆ ∆
= − = +
∆ ∆
∆
=
∫
( )
2
2
2
2 /
12 3
p p
q
m L m
P
L
= =
40
41. Signal-to-Quantization-Noise Ratio
2
2
Signal Power
Noise Power
3
.
s
q
s
p
P
SNR
P
L
P
m
= =
=
( )
( )
2
2
10 10 102 2
10 102
6
3 3
10 log 10 log 10 log 2
3
10 log 20 log 2
6 dB.
n
dB s s
p p
s
p
n
L
SNR P P
m m
P n
m
n
α
α
= × = × + × ÷ ÷ ÷ ÷
= × + × ÷ ÷
= +
1442443
144424443
41
42. SNR-Bandwidth Exchange
• More bits/sample for the same message results in
more quantization levels, less quantization step, less
quantization noise, higher SNR.
• On the other hand, more bits/sample results in
bandwidth expansion
• One added bit results in multiplying SNR by a factor of
4 (6 dB), but multiplying the transmission bandwidth by
a factor of (n+1)/n
nSNR
m
P
SNR dB
n
p
s
6)(;)2(3 2
+== α
42
43. • A signal of bandwidth 4 kHz is samples at Nyquist rate
and transmitted using PCM with uniform quantization. If
the number of quantization levels L is increased from 64
to 256, find the change in SNR and transmission
bandwidth.
– Number of bits/sample has been increased from 6 to 8.
– SNR improved by 12 dB (16 times)
– BT expanded by a factor of 1.33 (33% increase).
From 24 kHz to 32 kHz.
43
44. Non-Uniform Quantization
• There is a huge variation in voice signal level
from user to user, and for the same use from
call to call as well as within the call
(sometimes of the order of 1000:1)
• Uniform quantization provides same degree of
resolution for low and high values.
• Designing the step size for the low values
results in too many levels, and designing them
for the high values destroys the low values.
44
46. Compressors and Expanders
• It is practically more feasible to compress the signal
logarithmically then apply it to a uniform quantizer.
• A reciprocal process takes place at the receiver by an
expander.
• The compressor/expander system is called compander.
• There are two standard laws for companders, the µ-law
(North America and Japan) and the A-law (Europe and
rest of the world).
46
48. 48
Introduction
• Pulse modulation consists essentially of sampling analog
information signals and then converting those samples into
discrete pulses and transporting the pulses from a source to a
destination over a physical transmission medium.
• The four predominant methods of pulse modulation:
– pulse width modulation (PWM)
– pulse position modulation (PPM)
– pulse amplitude modulation (PAM)
– pulse code modulation (PCM).
50. 50
Pulse Width Modulation
• PWM is sometimes called pulse duration modulation
(PDM) or pulse length modulation (PLM), as the
width (active portion of the duty cycle) of a constant
amplitude pulse is varied proportional to the
amplitude of the analog signal at the time the signal
is sampled.
• The maximum analog signal amplitude produces the
widest pulse, and the minimum analog signal
amplitude produces the narrowest pulse. Note,
however, that all pulses have the same amplitude.
51. 51
Pulse Position Modulation
• With PPM, the position of a constant-width pulse
within a prescribed time slot is varied according to
the amplitude of the sample of the analog signal.
• The higher the amplitude of the sample, the farther
to the right the pulse is positioned within the
prescribed time slot. The highest amplitude sample
produces a pulse to the far right, and the lowest
amplitude sample produces a pulse to the far left.
52. 52
Pulse Amplitude Modulation
• With PAM, the amplitude of a constant width,
constant-position pulse is varied according to the
amplitude of the sample of the analog signal.
• The amplitude of a pulse coincides with the
amplitude of the analog signal.
• PAM waveforms resemble the original analog signal
more than the waveforms for PWM or PPM.
53. 53
Pulse Code Modulation
• With PCM, the analog signal is sampled and
then converted to a serial n-bit binary code
for transmission.
• Each code has the same number of bits and
requires the same length of time for
transmission
54. 54
Pulse Modulation
• PAM is used as an intermediate form of modulation
with PSK, QAM, and PCM, although it is seldom used
by itself.
• PWM and PPM are used in special-purpose
communications systems mainly for the military but
are seldom used for commercial digital transmission
systems.
• PCM is by far the most prevalent form of pulse
modulation and will be discussed in more detail.
55. 55
Pulse Code Modulation
• PCM is the preferred method of communications
within the public switched telephone network
because with PCM it is easy to combine digitized
voice and digital data into a single, high-speed digital
signal and propagate it over either metallic or optical
fiber cables.
• The term pulse code modulation is somewhat of a
misnomer, as it is not really a type of modulation but
rather a form of digitally coding analog signals.
56. 56
Pulse Code Modulation
• With PCM, the pulses are of fixed length and fixed
amplitude.
• PCM is a binary system where a pulse or lack of a
pulse within a prescribed time slot represents either
a logic 1 or a logic 0 condition.
• PWM, PPM, and PAM are digital but seldom binary,
as a pulse does not represent a single binary digit
(bit).
58. 58
PCM Sampling
• The function of a sampling circuit in a PCM transmitter is to
periodically sample the continually changing analog input
voltage and convert those samples to a series of constant-
amplitude pulses that can more easily be converted to binary
PCM code.
• For the ADC to accurately convert a voltage to a binary code,
the voltage must be relatively constant so that the ADC can
complete the conversion before the voltage level changes. If
not, the ADC would be continually attempting to follow the
changes and may never stabilize on any PCM code.
59. 59
PCM Sampling
• Essentially, there are two basic techniques used to
perform the sampling function
– natural sampling
– flat-top sampling
• Natural sampling is when tops of the sample pulses
retain their natural shape during the sample interval,
making it difficult for an ADC to convert the sample
to a PCM code.
62. 62
PCM Sampling
• The most common method used for sampling
voice signals in PCM systems is flat- top
sampling, which is accomplished in a sample-
and-hold circuit. The purpose of a sample-and-
hold circuit is to periodically sample the
continually changing analog input voltage and
convert those samples to a series of constant-
amplitude PAM voltage levels.
63. 63
PCM Sampling Rate
• The Nyquist sampling theorem establishes the
minimum sampling rate (fs) that can be used for a
given PCM system.
• For a sample to be reproduced accurately in a PCM
receiver, each cycle of the analog input signal (fa)
must be sampled at least twice.
• Consequently, the minimum sampling rate is equal
to twice the highest audio input frequency.
64. 64
PCM Sampling Rate
• If fs is less than two times fa an impairment
called alias or foldover distortion occurs.
• Mathematically, the min- imum Nyquist
sampling rate is:
fs ≥ 2fa
65. 65
PCM Sampling Rate
• A sample-and-hold circuit is a nonlinear device
(mixer) with two inputs: the sampling pulse and the
analog input signal. Consequently, nonlinear mixing
occurs between these two signals.
• The output includes the two original inputs (the
audio and the fundamental frequency of the
sampling pulse), their sum and difference
frequencies (fs ± fa), all the harmonics of fs and fa (2fs,
2fa, 3fs, 3fa, and so on), and their associated cross
products (2fs ± fa,3fs ± fa, and so on).
67. 67
Example 1
• For a PCM system with
a maximum audio
input frequency of 4
kHz, determine the
minimum sample rate
and the alias frequency
produced if a 5-kHz
audio signal were
allowed to enter the
sample-and-hold
circuit.
68. Differential Pulse Code Modulation (DPCM)
• In PCM we quantize the analog samples. Since the signal
varies over a large range of amplitudes, we generally need a
large number of levels (an hence bits).
• Note that neighboring samples are “close” to each other in
values.
• If we instead quantize the difference between successive
samples, we will be dealing with much smaller range of
values.
• This will results in either:
– Using less number of bits for the same SNR.
– Obtaining smaller SNR for the same number of bits.
• Quantization noise will be reduced by a factor of (mp/md)2
68
69. Block Diagram of DPCM
[ ] [ ] [ 1].d k x k x k= − −
ˆ ˆ[ ] [ ] [ 1].qx k d k x k= + −
69
70. Generalized DPCM
• We can get even a smaller range of values if we define the
difference as:
• The more previous samples included, the better the
approximation, the smaller the difference.
• The relation d[k] = x[k]- x[k-1] is a special case where the
previous sample is taken as a prediction of the current value.
+−+−+−=
−=
)3()2()1()(ˆ
,ofvaluespreviousfrompredictedbecan)(ˆ
][ˆ][][
321 kxakxakxakx
xkx
kxkxkd
70
76. Shannon-Nyquist's Sampling Theorem
• A sampled time signal must not contain
components at frequencies above half the
sampling rate (The so-called Nyquist
frequency)
• The highest frequency which can be
accurately represented is one-half of the
sampling rate
76
77. Range of Human Hearing
• 20 – 20,000 Hz
• We lose high frequency response with age
• Women generally have better response than
men
• To reproduce 20 kHz requires a sampling rate
of 40 kHz
– Below the Nyquist frequency we introduce
aliasing
77
78. Effect of Aliasing
• Fourier Theorem states that any waveform
can be reproduced by sine waves.
• Improperly sampled signals will have other
sine wave components.
78
79. Half the Nyquist Frequency
-1.5
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20 25
79
82. Key Parameters
• Sampling frequency
– 11.025kHZ or 22.05kHZ or 44.1kHZ
• Number of bits per sample
– 8 bits (256 levels) or 16 bits (65,536 levels)
82
83. Digital Voice Telephone Transmission
• Voice data for telephony purposes is limited to
frequencies less than 4,000 Hz.
• According to Nyquist, it would take 8,000 samples (2
times 4,000) to capture a 4,000 Hz signal perfectly.
• Generally, one byte is recorded per sample (256
levels). One byte is eight bits of binary data.
• (8 bits * 8,000 samples per second = 64K bps) over a
circuit.
83
84. T-1 Transmisson
• T carrier circuits are designed around this
requirement, since they are primarily designed to
carry analog voice signals that have been digitalized.
• For example, look at the DS-1 signal which passes
over a T-1 circuit. For DS-1 transmissions, each frame
contains 8 bits per channel and there are 24
channels. Also, 1 "framing bit" is required for each of
the 24 channel frames.
84
85. T-1 Transmissons
• (24 channels * 8 bits per channel) + 1 framing
bit = 193 bits per frame.
193 bits per frame * 8,000 "Nyquist" samples
= 1,544,000 bits per second.
• And it just so happens that the T-1 circuit is
1.544 Mbps.--not a coincidence. Each of the
24 channels in a T-1 circuit carries 64Kbps.
85
86. • DS0 - 64 kilobits per second
• ISDN - Two DS0 lines plus signaling (16 kilobytes
per second), or 128 kilobits per second
• T1 - 1.544 megabits per second (24 DS0 lines)
• T3 - 43.232 megabits per second (28 T1s)
• OC3 - 155 megabits per second (84 T1s)
• OC12 - 622 megabits per second (4 OC3s)
• OC48 - 2.5 gigabits per seconds (4 OC12s)
• OC192 - 9.6 gigabits per second (4 OC48s)
Standards
86
89. CD ROMS
• Sampling rate is 44.1 kHz
• Nyquist Theorem says that the highest
reproduced frequency is 22.05 kHz.
– Any frequency above 22.05 kHz will produce
aliasing
• A low pass filter is used to block frequencies
above 22.05 kHz.
89
90. Problems with D/A
• Imperfect low pass filters
• Ideally you want 0 dB attenuation at 20 kHz
going up to 90 dB at 22 kHz
– Very expensive
• Oversampling will help
– Sample at 8 X 20 kHz = 160 kHz
• Then the low pass filtering needs to be accomplished in
140 kHz not 2 kHz
90
91. • Finite word length
– Most systems today do 16 bit digitizing
– 65536 different levels
• The loudest sounds need room, so the normal
sounds don’t use the entire range
– Problems occur at the low levels where sounds are
represented by only one or two bits. High distortions
result.
• Dithering adds low level broadband noise
Problems with D/A
91
92. • Clock speed variation (Jitter)
Problems with D/A
92
93. 93
DELTA MODULATION
• Delta modulation uses a single-bit PCM code to achieve digital
transmission of analog signals.
• With conventional PCM, each code is a binary representation
of both the sign and the magnitude of a particular sample.
Therefore, multiple-bit codes are required to represent the
many values that the sample can be.
• With delta modulation, rather than transmit a coded
representation of the sample, only a single bit is transmitted,
which simply indicates whether that sample is larger or
smaller than the previous sample.
94. 94
DELTA MODULATION
• The algorithm for a delta modulation system is
quite simple.
• If the current sample is smaller than the
previous sample, a logic 0 is transmitted.
• If the current sample is larger than the
previous sample, a logic 1 is transmitted.
96. 96
DELTA MODULATION
Transmitter
• The input analog is sampled and converted to a PAM signal,
which is compared with the output of the DAC.
• The output of the DAC is a voltage equal to the regenerated
magnitude of the previous sample, which was stored in the
up-down counter as a binary number.
• The up-down counter is incremented or decremented
depending on whether the previous sample is larger or
smaller than the current sample.
97. 97
DELTA MODULATION
Transmitter
• The up-down counter is clocked at a rate equal to the sample
rate. Therefore, the up-down counter is updated after each
comparison.
• Initially, the up-down counter is zeroed, and the DAC is
outputting 0 V. The first sample is taken, converted to a PAM
signal, and compared with zero volts.
• The output of the comparator is a logic 1 condition (+ V),
indicating that the current sample is larger in amplitude than
the previous sample. On the next clock pulse, the up--down
counter is incremented to a count of 1.
100. 100
Problems with DM
• Slope overload - when the analog input signal
changes at a faster rate than the DAC can maintain.
The slope of the analog signal is greater than the
delta modulator can maintain and is called slope
overload.
• Increasing the clock frequency reduces the
probability of slope overload occurring. Another way
to prevent slope overload is to increase the
magnitude of the minimum step size.
102. 102
Problems with DM
• Granular noise. Figure 10-24 contrasts the original
and reconstructed signals associated with a delta
modulation system. It can be seen that when the
original analog input signal has a relatively constant
amplitude, the reconstructed signal has variations
that were not present in the original signal. This is
called granular noise. Granular noise in delta
modulation is analogous to quantization noise in
conventional PCM.
103. 103
Problems with DM
• Granular noise can be reduced by decreasing the step size.
Therefore, to reduce the granular noise, a small resolution is
needed, and to reduce the possibility of slope overload
occurring, a large resolution is required. Obviously, a
compromise is necessary.
•
• Granular noise is more prevalent in analog signals that have
gradual slopes and whose amplitudes vary only a small
amount. Slope overload is more prevalent in analog signals
that have steep slopes or whose amplitudes vary rapidly.
105. 105
Adaptive DM
• Adaptive delta modulation is a delta
modulation system where the step size of the
DAC is automatically varied, depending on the
amplitude characteristics of the analog input
signal.
• Figure 10-25 shows how an adaptive delta
modulator works.
107. 107
Adaptive DM
• When the output of the transmitter is a string of consecutive
Is or 0s, this indicates that the slope of the DAC output is less
than the slope of the analog signal in either the positive or
the negative direction. Essentially, the DAC has lost track of
exactly where the analog samples are, and the possibility of
slope overload occurring is high. With an adaptive delta
modulator, after a predetermined number of consecutive 1s
or 0s, the step size is automatically increased. After the next
sample, if the DAC output amplitude is still below the sample
amplitude, the next step is increased even further until
eventually the DAC catches up with the analog signal.
108. 108
Adaptive DM
• When an alternative sequence of 1s and 0s is occurring, this
indicates that the possibility of granular noise occurring is
high. Consequently, the DAC will automatically revert to its
minimum step size and, thus, reduce the magnitude of the
noise error.
• A common algorithm for an adaptive delta modulator is when
three consecutive 1s or 0s occur, the step size of the DAC is
increased or decreased by a factor of 1.5.
• Various other algorithms may be used for adaptive delta
modulators, depending on particular system requirements
109. 109
Differential DM
• In a typical PCM-encoded speech waveform, there are often
successive samples taken in which there is little difference
between the amplitudes of the two samples.
• This necessitates transmitting several identical PCM codes,
which is redundant.
• Differential pulse code modulation (DPCM) is designed
specifically to take advantage of the sample-to-sample
redundancies in typical speech waveforms.
110. 110
Differential DM
• With DPCM, the difference in the amplitude of
two successive samples is transmitted rather
than the actual sample. Because the range of
sample differences is typically less than the
range of individual samples, fewer bits are
required for DPCM than conventional PCM.
111. Modulation of Digital Data
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Quadrature Amplitude Modulation
Bit/Baud Comparison
111
114. Bit rate is the number of bits per
second. Baud rate is the number of
signal units per second. Baud rate is
less than or equal to the bit rate.
Note:Note:
114
115. Example 1Example 1
An analog signal carries 4 bits in each signal unit. If 1000
signal units are sent per second, find the baud rate and the
bit rate
SolutionSolution
Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps
115
116. Example 2Example 2
The bit rate of a signal is 3000. If each signal unit carries
6 bits, what is the baud rate?
SolutionSolution
Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s
116
119. Example 3Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is half-
duplex.
SolutionSolution
In ASK the baud rate and bit rate are the same. The baud
rate is therefore 2000. An ASK signal requires a
minimum bandwidth equal to its baud rate. Therefore,
the minimum bandwidth is 2000 Hz.
119
120. Example 4Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what
are the baud rate and bit rate?
SolutionSolution
In ASK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But because the baud
rate and the bit rate are also the same for ASK, the bit
rate is 5000 bps.
120
121. Example 5Example 5
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz),
draw the full-duplex ASK diagram of the system. Find
the carriers and the bandwidths in each direction. Assume
there is no gap between the bands in the two directions.
SolutionSolution
For full-duplex ASK, the bandwidth for each direction is
BW = 10000 / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of
each band (see Fig. 5.5).
fc (forward) = 1000 + 5000/2 = 3500 Hz
fc (backward) = 11000 – 5000/2 = 8500 Hz
121
125. Example 6Example 6
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.
SolutionSolution
For FSK
BW = baud rate + fBW = baud rate + fc1c1 −− ffc0c0
BW = bit rate + fc1BW = bit rate + fc1 −− fc0 = 2000 + 3000 = 5000 Hzfc0 = 2000 + 3000 = 5000 Hz
125
126. Example 7Example 7
Find the maximum bit rates for an FSK signal if the
bandwidth of the medium is 12,000 Hz and the difference
between the two carriers is 2000 Hz. Transmission is in
full-duplex mode.
SolutionSolution
Because the transmission is full duplex, only 6000 Hz is
allocated for each direction.
BW = baud rate + fc1BW = baud rate + fc1 −− fc0fc0
Baud rate = BWBaud rate = BW −− (fc1(fc1 −− fc0 ) = 6000fc0 ) = 6000 −− 2000 = 40002000 = 4000
But because the baud rate is the same as the bit rate, the
bit rate is 4000 bps.
126
133. Example 8Example 8
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.
SolutionSolution
For PSK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But in 8-PSK the bit
rate is 3 times the baud rate, so the bit rate is 15,000 bps.
133
134. Example 9Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
SolutionSolution
For PSK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But in 8-PSK the bit
rate is 3 times the baud rate, so the bit rate is 15,000 bps.
134
135. 135
Time-division multiplexing (TDM) can be used to combine PAM
or PCM signals
2 total number of pulses / second
(With raised cos-pulses TDM baseband
bandwidth at least / 2 )
s
T
r Mf M W
B r MW
≥ = =
≥ =
136. 136
A TDM realization
• TDM systems are critical in timing
• Timing can be arranged by
– bit-stuffing
– marker pulses
– pilot tones
– utilizing statistical properties
of TDM signals
Analog switch matrix
137. 137
• Analog sources must be sampled at least at the rates: 1,3: 4kHz
and 2: 8 kHz
• Hence PAM samples generated at the rate of 2x4+8=16 kHz
• After 4 bit-digitization data stream has the rate of 4x16=64
kbit/s
• For the digital sources pulse stuffing (clock rate shift/jitter
compensation) increases the source rates to 8 kbit/s yielding
output rate of 8x8 = 64 kbit/s
• Buffers enable constant rate for the second rotating switch
8 Digital sources
3 Analog sources
*W Stallings: Data and Computer
Communications
138. 138
PCM systems and digital time division
multiplexing (TDM)
In digital multiplexing several messages are transmitted via
same physical channel. For multiplexing 64 kbit/s channels in
digital exchanges following three methods are popular:
PDH (plesio-synchronous digital hierarchy) (the dominant
method today) (‘50-’60, G.702)
SONET (synchronous optical network) (‘85)
SDH (synchronous digital hierarchy) (CCITT ‘88)
European PCM frame
32 time slots x 8 bits x 8000 Hz = 2048 kbit/s
frame synchronization slot
signaling or traffic
traffic
125 µs
PDH E-1 frame
139. 139
PCM hierarchy in PDH
139264 kbit/s139264 kbit/s
34368 kbit/s34368 kbit/s
8448 kbit/s8448 kbit/s
2048 kbit/s2048 kbit/s
64 kbit/s64 kbit/s
European hierarchy
x4
...
x4
x4
x32
139264 kbit/s139264 kbit/s
44736 kbit/s44736 kbit/s
6312 kbit/s6312 kbit/s
1544 kbit/s1544 kbit/s
64 kbit/s64 kbit/s
USA hierarchy
x3
...
x7
x4
x24
If one wishes to disassemble a tributary from the main
flow the main flow must be demultiplexed step by step to
the desired main flow level in PDH.
140. T1 and E1 Digital transmission systems
In PSTN two PCM systems dominate:
T1, developed by Bell Laboratories, used in USA
E1, developed by CEPT* used in most of other countries
In both data streams divided in frames of 8000 frames/sec
In T1 (E1)
24 (32) times slots and a framing (F) bit serves 24 channels
F bit (frame sync) repeats every 12 frames
Frame length: 1+ 8x24=193 bits
Rate 193x8000 bits/second=1544 kb/s
In E1 TS 0 holds a synchronization pattern and TS 16 holds
signaling information
An E1 frame has 32x8=256 bits and its rate is
8000x256=2048 kb/s
140
141. 141
PCM-method summarized
• Analog speech signal is applied into a LP-filter restricting its
bandwidth into 3.4 kHz
• Sampling circuit forms a PAM pulse train having rate of 8 kHz
• Samples are quantized into 256 levels that requires a 8 bit-word
for each sample (28
=256).
• Thus a telephone signal requires 8x8 kHz = 64 kHz bandwidth
• The samples are line coded by using the HDB-3 scheme to
alleviate synchronization problems at the receiver
• Usually one transmits several channels simultaneously
following SDH hierarchy (as 30 pcs)
• Transmission link can be an optical fiber, radio link or an
electrical cable
• At the receiver the PAM signal is first reconstructed where after
it is lowpass filtered to yield the original-kind, analog signal
